• poj 2955 Brackets (区间dp 括号匹配)


    Description

    We give the following inductive definition of a “regular brackets” sequence:

    • the empty sequence is a regular brackets sequence,
    • if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
    • if a and b are regular brackets sequences, then ab is a regular brackets sequence.
    • no other sequence is a regular brackets sequence

    For instance, all of the following character sequences are regular brackets sequences:

    (), [], (()), ()[], ()[()]

    while the following character sequences are not:

    (, ], )(, ([)], ([(]

    Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1i2, …, im where 1 ≤ i1i2 < … < im ≤ nai1ai2 … aim is a regular brackets sequence.

    Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

    Input

    The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters ()[, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

    Output

    For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

    Sample Input

    ((()))

    ()()()

    ([]])

    )[)(

    ([][][)

    end

    Sample Output

    6

    6

    4

    0

    6

    题意:现在已知一个由'()''[]'组成的括号序列 问其中括号匹配数最大的子序列个数

    思路:

    这里的状态转移是以一个if为基础的,如果s[i]与s[j]匹配,那么明显的dp[i][j] = dp[i+1][j-1];然后在这个基础上枚举分割点k.

    状态转移方程:dp[i][j]表示第i~j个字符间的最大匹配对数。

    if(s[i] 与 s[j]匹配) dp[i][j] = d[[i+1][j-1] +1;

    dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]);

    最后乘2即可

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<string>
    #include<vector>
    #include<stack>
    #include<bitset>
    #include<cstdlib>
    #include<cmath>
    #include<set>
    #include<list>
    #include<deque>
    #include<map>
    #include<queue>
    #define ll long long int
    using namespace std;
    inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
    int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
    int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
    const int inf=0x3f3f3f3f;
    const ll mod=1e9+7;
    int dp[107][107];
    string s;
    int jug(int i,int j){
        return (s[i]=='['&&s[j]==']')||(s[i]=='('&&s[j]==')');
    }
    int main(){
        ios::sync_with_stdio(false);
        
        while(cin>>s){
            if(s=="end") break;
            memset(dp,0,sizeof(dp));
            int n=s.length();
            for(int len=2;len<=n;len++){
                for(int i=1;i+len<=n+1;i++){
                    int j=i+len-1;
                    dp[i][j]=dp[i+1][j-1]+jug(i-1,j-1);
                    for(int k=i;k<j;k++)
                        dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
            cout<<2*dp[1][n]<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wmj6/p/10704805.html
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